Mathematics

$\displaystyle\int {\dfrac{{dx}}{{\sqrt {x + 1 + } \,\sqrt x }}}$

SOLUTION
$I=\displaystyle\int \dfrac{1}{\sqrt{x+1}+\sqrt{x}}dx$
$=\displaystyle\int \dfrac{1}{\sqrt{x+1}+\sqrt{x}}\times \dfrac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x+1}-\sqrt{x}}$
$=\displaystyle\int \dfrac{\sqrt{x+1}-\sqrt{x}}{x+1-x}dx$
$=\displaystyle\int (\sqrt{x+1})dx-\int \sqrt{x}dx$
$=\dfrac{2}{3}(x+1)^{3/2}-\dfrac{2}{3}x^{3/2}+c$
$=\dfrac{2}{3}[(x+1)^{3/2}-x^{3/2}]+c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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